# 선형대수응용_HW_W12

## 566 days 전, jongha4886 작성

x1 = vector([1,-2,-1,-3,2,3,0]) x2 = vector([3,-1,1,2,0,-3,-2]) x3 = vector([-4,-2,1,-1,2,-3,0]) x4 = vector([1,2,0,-2,3,-1,-3]) x5 = vector([2,-1,0,3,-2,1,-3]) x6 = vector([1,0,2,1,-1,2,3]) x7 = vector([-1,0,1,2,2,3,0]) A = matrix([x1,x2,x3,x4,x5,x6,x7]) if A.det() != 0: print "A의 모든 행벡터는 일차 독립이며 R^7을 생성합니다." print "따라서 주어진 벡터는 R^7의 기저입니다." B = A.gram_schmidt()[0] B = matrix([B.row(i) / B.row(i).norm() for i in range(0,7)]) if (B*B.transpose() == identity_matrix(7)) & (B.transpose()*B == identity_matrix(7)): print "다음 B의 행벡터들이 R^7의 정규직교기저입니다." print print B
 A의 모든 행벡터는 일차 독립이며 R^7을 생성합니다. 따라서 주어진 벡터는 R^7의 기저입니다. 다음 B의 행벡터들이 R^7의 정규직교기저입니다. [ 1/14*sqrt(7) -1/7*sqrt(7) -1/14*sqrt(7) -3/14*sqrt(7) 1/7*sqrt(7) 3/14*sqrt(7) 0] [ 95/1326*sqrt(663/7) -25/663*sqrt(663/7) 1/78*sqrt(663/7) 23/1326*sqrt(663/7) 11/663*sqrt(663/7) -1/26*sqrt(663/7) -28/663*sqrt(663/7)] [ -2279/22709*sqrt(22709/663) -1627/22709*sqrt(22709/663) 646/22709*sqrt(22709/663) -803/22709*sqrt(22709/663) 1538/22709*sqrt(22709/663) -1938/22709*sqrt(22709/663) -178/22709*sqrt(22709/663)] [ 1508/524621*sqrt(524621/22709) 77215/524621*sqrt(524621/22709) -3078/524621*sqrt(524621/22709) -33577/524621*sqrt(524621/22709) 44769/524621*sqrt(524621/22709) -13475/524621*sqrt(524621/22709) -51249/524621*sqrt(524621/22709)] [ -317732/1462919*sqrt(1462919/524621) -344115/2925838*sqrt(1462919/524621) -45865/1462919*sqrt(1462919/524621) 529915/2925838*sqrt(1462919/524621) -145399/1462919*sqrt(1462919/524621) 675615/2925838*sqrt(1462919/524621) -1310511/2925838*sqrt(1462919/524621)] [ 16381/1273669*sqrt(2547338/1462919) 604911/10189352*sqrt(2547338/1462919) 846679/1273669*sqrt(2547338/1462919) 1353693/10189352*sqrt(2547338/1462919) 1028445/5094676*sqrt(2547338/1462919) 2599835/10189352*sqrt(2547338/1462919) 734773/10189352*sqrt(2547338/1462919)] [ -54*sqrt(1/2547338) 247/4*sqrt(1/2547338) -614*sqrt(1/2547338) 3989/4*sqrt(1/2547338) 2037/2*sqrt(1/2547338) 691/4*sqrt(1/2547338) 1277/4*sqrt(1/2547338)] A의 모든 행벡터는 일차 독립이며 R^7을 생성합니다. 따라서 주어진 벡터는 R^7의 기저입니다. 다음 B의 행벡터들이 R^7의 정규직교기저입니다. [ 1/14*sqrt(7) -1/7*sqrt(7) -1/14*sqrt(7) -3/14*sqrt(7) 1/7*sqrt(7) 3/14*sqrt(7) 0] [ 95/1326*sqrt(663/7) -25/663*sqrt(663/7) 1/78*sqrt(663/7) 23/1326*sqrt(663/7) 11/663*sqrt(663/7) -1/26*sqrt(663/7) -28/663*sqrt(663/7)] [ -2279/22709*sqrt(22709/663) -1627/22709*sqrt(22709/663) 646/22709*sqrt(22709/663) -803/22709*sqrt(22709/663) 1538/22709*sqrt(22709/663) -1938/22709*sqrt(22709/663) -178/22709*sqrt(22709/663)] [ 1508/524621*sqrt(524621/22709) 77215/524621*sqrt(524621/22709) -3078/524621*sqrt(524621/22709) -33577/524621*sqrt(524621/22709) 44769/524621*sqrt(524621/22709) -13475/524621*sqrt(524621/22709) -51249/524621*sqrt(524621/22709)] [ -317732/1462919*sqrt(1462919/524621) -344115/2925838*sqrt(1462919/524621) -45865/1462919*sqrt(1462919/524621) 529915/2925838*sqrt(1462919/524621) -145399/1462919*sqrt(1462919/524621) 675615/2925838*sqrt(1462919/524621) -1310511/2925838*sqrt(1462919/524621)] [ 16381/1273669*sqrt(2547338/1462919) 604911/10189352*sqrt(2547338/1462919) 846679/1273669*sqrt(2547338/1462919) 1353693/10189352*sqrt(2547338/1462919) 1028445/5094676*sqrt(2547338/1462919) 2599835/10189352*sqrt(2547338/1462919) 734773/10189352*sqrt(2547338/1462919)] [ -54*sqrt(1/2547338) 247/4*sqrt(1/2547338) -614*sqrt(1/2547338) 3989/4*sqrt(1/2547338) 2037/2*sqrt(1/2547338) 691/4*sqrt(1/2547338) 1277/4*sqrt(1/2547338)]